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BIO 304 Genetics (Fall 2003)
transformation
Southern blotting
nonsense mutation
missense mutation
electrophoresis
transduction
reverse transcriptase
transposable element
dicentric
lysogeny
Northern blotting
nondisjunction
frameshift
conjugation
sister chromatids
complementation
thymine dimerization
temperate phage
allopolyploid
glycosylase
monoploid
auxotroph
Exam #2
conditional mutation
hybridization
translocation
alkyltransferase
nitrocellulose
silent mutation
heterokaryon
plasmid
transversion
complementation
tautomerization
autopolyploid
Prozac
reversion
RFLP
nonsense suppressor
intercalating agent
prophage
clone
catalase
PCR
library
Name_______KEY_______________
SSN____________________________
penetrance
epistasis
amplification
vector
VNTR
inverted repeats
bacteriophage
probe
episome
histone
inversion
ligase
expression vector
cDNA
biotechnology
transition
polyploid
loss-of-function
synteny
lytic
pleiotropy
stem cells
expressivity
co-dominance
transposase
long terminal repeats
ampicillin
polymerase
duplication
restriction enzyme
suppression
dideoxy sequencing
F factor
polymorphism
independent assortment
two-hybrid
heteroduplex DNA
tautomeric shift
aneuploid
gain-of-function
replicative transposition
conservative transposition
homeologous
crossover
Use one of the above terms to best complete each sentence #1-15 below. (2 pts. each)
1. _____cDNA_______ is a DNA copy of an RNA molecule.
2. ___reverse transcriptase__ is an RNA-dependent DNA polymerase.
3. Knockout mice are created by replacing a normal gene segment with a modified segment
within embryonic _stem cells_______, then using the latter to create a chimeric embryo.
4. __transformation____ is a method of DNA transfer in bacteria in which environmental DNA is
taken up through the cell wall and membrane of the cell.
5. A ___library________ is a collection of DNA clones that represent the genome or RNA
population of an organism or tissue.
6. A restriction site difference that is polymorphic within a population can be used to carry out
_____RFLP___________ mapping of neighboring genes.
7. ___histone___________ genes commonly exist in tandem repetitive arrays in eukayote
genomes.
8. ____transposase________ is a transposon-encoded protein that mediates movement of the
transposon.
9. ____two-hybrid________ analysis detects physical interactions between two proteins in yeast
cells, by fusing each protein to a separate domain of the yeast GAL4 regulatory protein and
studying the ability of the combination to activate a UAS-lacZ gene.
10. _____VNTR_________________ sequences are one- to five-kilobase sequences consisting of
variable numbers of repeat units, each unit being 15 to 100 nucleotides in length.
11. _____catalase___________ is an enzyme that prevents DNA damage by reducing reactive
oxygen species.
12. A type of DNA damage caused typically by UV light is ___thymine dimerization_____.
13. A karyotype that has an extra set of chromosomes is a type of ____polyploid____ genome.
14. Retrotransposons have sequences at the ends that are best described as ___long terminal
repeats______.
15. ___translocation_____________ is a rearrangement that moves a segment of DNA to a new
chromosome.
*
*
*
*
*
*
*
*
*
*
*
*
*
16. Wild-type tigers are orange and black with white stripes, have eyes that glow in the dark and
have bad breath. A mutant strain of tigers is developed in which animals are albino (a),
have glowless eyes (g) and sweet breath (b), each determined by a single gene. Truebreeding wild-type females were crossed to males having all three mutant traits, producing
an F1 of entirely wild-type animals. F1 females were testcrossed to a g b males, producing an
astonishing number of progeny:
Phenotype
Number
a+ g+ b+
a g b
a+ g b
a g+ b+
a+ g b+
a g+ b
a g b+
429
433
55
13
15
2
----1000
parental types
53 SCO I (a-b) - 108
SCO II (g-b) - 28
DCO - 2
a. Draw a map of these three genes, indicating their order and the map distances between
them (10 pts.)
a
108 + 2 = 110
28 + 2 = 30
110 ÷ 1000 = .11
30 ÷ 1000 = .03
11 m.u.
b
3 m.u.
g
b. Calculate the coefficient of coincidence and the interference value for these results.
(3 pts.)
COC = observed DCO ÷ expected DCO = 2 ÷ [.03 X .11] = 2 / 3.3 = 0.61
interference = 1- COC = 1 – 0.61 = 0.39
17. A cross is made between an E. coli Hfr strain that is a+ b+ d+ and an F- strain that is a- b- d-.
Interrupted mating studies showed that b+ enters the recipient strain last. The b+
recombinants were then tested for the presence of the a and d alleles, producing the
following results:
a+ b+ d +
a- b+ d +
a+ b+ d a- b+ d -
326
2
14
58
-----400
parental type
DCO
SCO I (a-d)
SCO II (a-b)
Draw a map of these genes, showing the gene order and frequencies of recombination
between each pair. (6 pts.)
14 + 2 = 16
16 ÷ 400 = .04
d
4%
58 + 2 = 60
60 ÷ 400 = 0.15
a
15%
b
18. DNA studies were performed on a large family that shows an autosomal dominant disease of
late onset. A DNA sample from each member is digested with TaqI and subjected to
Southern blot analysis, using a particular unique sequence probe:
Pedigree:
Migration
5 kb
Autoradiogram:
3 kb
2kb
a. What is the relationship between the restriction pattern observed with this probe and the
gene for the disease? (7 pts.)
The 3+2 kb morph is linked to the disease gene; the 5 kb morph is linked to its wild-type allele.
This correlation applies in 8 of 9 progeny.
b. How do you explain the pattern seen in the rightmost son? (3 pts.)
A recombination event occurred between the two sites in the father, producing the disease
allele in cis with the 5 kb morph.
19. Using dideoxy nucleotide DNA sequencing, you observe the following autoradiogram:
What is the 5’-3’ nucleotide sequence of the template strand that you have sequenced?
(6 pts.)
5’ A T G C A 3’
20. Geneticists working on the Human Genome Project, focusing on one band of chromosome
11, had a collection of eight cloned fragments (clones 1-8). They also determined that seven
sequence-tagged sites (STSa-g) were distributed among these clones. The STS distribution on
the various clones was as follows:
Clone
STS content
1
2
3
4
5
6
7
8
a, b, e, f
a, b
f
a, c, f
c, d
e
g
e, g
Arrange these clones into a contig, showing the regions of overlap and positions of STSs.
(8 pts.)
STS:
d
c
1
5
f
a
b
e
f
a
b
e
2
a
b
3
f
4
c
f
d
c
8
e
6
e
g
g
a
7
g
21. For each of the following aberrant sex chromosome configurations, a nondisjunction event is
responsible. Indicate during which meiotic division the nondisjunction could have occurred
assuming first that it occurred in the mother, then assuming it occurred in the father. Your
possible answers for each are: meiosis I, meiosis II, either, or not possible. (12 pts.)
Offspring
Meiotic division if nondisjunction
occurred in mother
Meiotic division if nondisjunction
occurred in father
XXY
either
Meiosis I
XYY
not possible
Meiosis II
XXX
either
Meiosis II
22. What is meant by missense, nonsense, and synonymous mutations? Which of the three is
most likely to disrupt gene function and why? (6 pts.)
Missense mutations are changes that cause the substitution of one amino acid for another in the
encoded protein.
Nonsense mutations are changes that cause the substitution of a stop codon for an amino acid
in the encoded protein.
Synonymous mutations are changes in the nucleic acid sequence in the coding region of a
gene that do not cause a change in the encoded protein.
Nonsense mutations are most likely to disrupt gene function because they cause the truncation
of the encoded protein, deleting all amino acids that follow the mutation in the coding
sequence.
23. Indicate with an “X” which of the following enzymes act in each of the listed E. coli mutation
repair systems. Enzymes may be active in none or more than one of the listed systems. (6 pts)
Excision repair
Recombinational
repair
(either OK)
Mismatch repair
UV photodimer
splitting
DNA polymerase I X
X
Photolyase
X
DNA ligase
X
(either OK)
X
Glycoslylase
X
RecA
X
(One point for each correct X, minus one point for each incorrect X, sum not < 0 or >7)